Use the standard series for $\ln(1+x)$ and $\ln(1-x)$ to find the first three non-zero terms in the series for $\ln\left(\dfrac{1+x}{1-x}\right)$. By choosing a suitable value for $x$ in your answer for part (i), find $\displaystyle\sum_{r=0}^{\infty}\frac{1}{(2r+1)4^r}.$ Given that $\displaystyle\sum_{r=1}^n r^2=\frac{n}{6}(n+1)(2n+1),$ find, in terms of $a$, $\displaystyle\sum_{r=37}^a (2r)^2,\quad\text{where $a$ is an integer and }a\ge 37.$

1. Express the logarithm of a quotient as the difference of logarithms and state the standard series for each up to the $x^5$ term.

$\ln\left(\frac{1+x}{1-x}\right) = \ln(1+x) - \ln(1-x)$
Using the standard series expansions:
$\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5} - \dots$
$\ln(1-x) = -x - \frac{x^2}{2} - \frac{x^3}{3} - \frac{x^4}{4} - \frac{x^5}{5} - \dots$

2. Subtract the series to obtain the first three non-zero terms.

$\ln\left(\frac{1+x}{1-x}\right) = \left( x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5} \right) - \left( -x - \frac{x^2}{2} - \frac{x^3}{3} - \frac{x^4}{4} - \frac{x^5}{5} \right) + \dots$
$= 2x + \frac{2x^3}{3} + \frac{2x^5}{5} + \dots$

1. Write out the first few terms of the given infinite sum and manipulate them to match the form of the expansion from part (i).

From part (i), we established the expansion:
$\ln\left(\dfrac{1+x}{1-x}\right) = 2x + \dfrac{2x^3}{3} + \dfrac{2x^5}{5} + \dots$

Expanding the first few terms of the target sum:
$\sum_{r=0}^{\infty}\frac{1}{(2r+1)4^r} = 1 + \frac{1}{3(4)} + \frac{1}{5(16)} + \dots$

Rewriting these terms to match the form of the Maclaurin expansion $2x + \frac{2x^3}{3} + \frac{2x^5}{5} + \dots$:
$1 + \frac{1}{12} + \frac{1}{80} + \dots = 2\left(\frac{1}{2}\right) + \frac{2}{3}\left(\frac{1}{2}\right)^3 + \frac{2}{5}\left(\frac{1}{2}\right)^5 + \dots$

2. Identify the value of $x$ by comparing the series, and evaluate the logarithmic expression to find the exact sum.

Comparing the terms with the expansion, it is clear that $x = \dfrac{1}{2}$.

Substitute $x = \dfrac{1}{2}$ into the original logarithmic function:
$\text{Sum} = \ln\left(\dfrac{1 + \frac{1}{2}}{1 - \frac{1}{2}}\right)$
$\text{Sum} = \ln\left(\dfrac{\frac{3}{2}}{\frac{1}{2}}\right)$
$\text{Sum} = \ln 3$

1. Factor out the constant $2^2=4$ from the summation.

$\sum_{r=37}^a (2r)^2 = \sum_{r=37}^a 4r^2 = 4 \sum_{r=37}^a r^2$

2. Express the summation from $r=37$ to $a$ as the difference of two summations starting from $r=1$.

$4 \sum_{r=37}^a r^2 = 4 \left( \sum_{r=1}^a r^2 - \sum_{r=1}^{36} r^2 \right)$

3. Substitute the given formula for both sums and simplify to get the final expression.

$4 \left[ \frac{a}{6}(a+1)(2a+1) - \frac{36}{6}(36+1)(2(36)+1) \right]$
$= 4 \left[ \frac{a}{6}(a+1)(2a+1) - 6(37)(73) \right]$
$= \frac{2}{3}a(a+1)(2a+1) - 4(16206)$
$= \frac{2}{3}a(a+1)(2a+1) - 64824$